SATHEE: Application of Derivatives 2 Question 27

Application of Derivatives 2 Question 27

####27. Using the relation $2 (1 - \cos x) < x^{2}, x \neq 0$ or prove that $\sin (\tan x) \geq x, \forall x \in [0, π / 4]$ .

(2003, 4M)

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Answer:

Correct Answer: 27. (c)

Solution:

$\begin{aligned} f (x) & = \sin (\tan x) - x \\ f^{'} (x) & = \cos (\tan x) \cdot \sec^{2} x - 1 \\ = \cos (\tan x) (1 + \tan^{2} x) - 1 \end{aligned}$

$= \tan^{2} x \cos (\tan x) + \cos (\tan x) - 1$

$> \tan^{2} x \cos (\tan x) - \frac{\tan^{2} x}{2}$

$∵ 2 (1 - \cos x) < x^{2}, x \neq 0 \Rightarrow \cos x > 1 - \frac{x^{2}}{2}$

$\Rightarrow \cos (\tan x) > 1 - \frac{\tan^{2} x}{2}$

$f^{'} (x) > \tan^{2} x \cos (\tan x) - \frac{1}{2}$

$> \tan^{2} x [\cos (\tan x) - \cos (π / 3)] > 0$

$\Rightarrow f (x)$ is increasing function, for all $x \in [0, π / 4]$

$\begin{aligned} f (0) = 0 \Rightarrow & f (x) \geq 0, for all x \in [0, π / 4] \\ \sin (\tan x) \geq x \end{aligned}$

$\Rightarrow$

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Application of Derivatives 2 Question 27

Answer:

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